Question

Solve the following questions​

Answer 1

Start with

$\dfrac{2x-1}{2x+1}-\dfrac{2x+1}{2x-1}$

Observe that this implies that the domain is

$2x+1,\ 2x-1\neq 0 \implies x \neq \pm\dfrac{1}{2}$

The least common denominator is

$\dfrac{2x-1}{2x+1}-\dfrac{2x+1}{2x-1} = \dfrac{(2x-1)^2-(2x+1)^2}{(2x+1)(2x-1)}=\dfrac{-8x}{4x^2-1}$

In the right hand side, we have

$-2\dfrac{2}{3}=-\left(2+\dfrac{2}[3}\right)=-\left(\dfrac{6}{3}+\dfrac{2}{3}\right)=-\dfrac{8}{3}$

So, the equation becomes

$-\dfrac{8x}{4x^2-1}=-\dfrac{8}{3} \iff \dfrac{x}{4x^2-1}=\dfrac{1}{3}$

Multiply both sides by 3:

$\dfrac{3x}{4x^2-1}=1 \iff 3x=4x^2-1 \iff 4x^2-3x-1=0$

This equation has solutions

$x=-\dfrac{1}{4},\quad x=1$

Which are compliant with the domain of the equation